\(2x - 3y = 4\)
\(
x+ 3y = 2\)
\(\Leftrightarrow\) \(3x = 6\)
Sửa đề: Tìm \(x,y\in Z\)
\(\Leftrightarrow2x=4+3y\)
\(\Leftrightarrow x=\dfrac{4+3y}{2}\)
\(\Leftrightarrow x=\dfrac{4}{2}+\dfrac{2y}{2}+\dfrac{y}{2}\)
\(\Leftrightarrow x=2+y+\dfrac{y}{2}\)
Để x,y nguyên thì \(\dfrac{y}{2}\) nguyên
Đặt \(\dfrac{y}{2}=k\)
\(\Leftrightarrow y=2k\)
\(\Rightarrow x=2+2k+k\)
\(\Leftrightarrow x=2+3k\)
Vậy nghiệm nguyên của pt \(\left\{{}\begin{matrix}x=2+3k\\y=2k;k=\dfrac{y}{2}\in Z\end{matrix}\right.\)
\(2x-3y=4\)
\(2x-3y+3y=4+3y\)
\(2x=4+2y\)
\(\dfrac{2x}{2}=\dfrac{4}{2}+\dfrac{3y}{2}\)
\(x=\dfrac{4+3y}{2}\)
